Kaplunovsky - practice 09

# Kaplunovsky - practice 09 - practice 09 – – Due 1:00 pm...

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Unformatted text preview: practice 09 – – Due: Mar 27 2007, 1:00 pm 1 Question 1, chap 11, sect 3. part 1 of 1 8 points A disc of mass m moves horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 7 m . The second disc is moving horizon- tally to the right with speed v 2 at the moment of impact. m 7 m v v 2 before The two discs stick together upon impact. 8 m v f after What is the speed of the composite body immediately after the collision? 1. v f = 2 7 v 2. v f = 7 27 v 3. v f = 5 8 v 4. v f = 9 16 v correct 5. v f = 1 5 v 6. v f = 11 32 v 7. v f = 3 5 v 8. v f = 13 40 v 9. v f = 11 35 v 10. None of these are correct. Explanation: The total momentum of the system is con- served because there is no exterior force. So what we have is ( m + 7 m ) v f = mv + 7 m v 2 8 mv f = mv parenleftbigg 1 + 7 2 parenrightbigg 8 v f = parenleftbigg 2 2 + 7 2 parenrightbigg v 8 v f = 9 2 v v f = 9 (2) (8) v = 9 16 v , so = 9 16 v . Question 2, chap 10, sect 99. part 1 of 3 8 points Two small spheres of mass 478 g and 611 g are suspended from the ceiling at the same point by massless strings of equal length 10 . 2 m. The lighter sphere is pulled aside through an angle of 49 ◦ from the vertical and let go. The acceleration of gravity is 9 . 8 m / s 2 . 9 . 8 m / s 2 Before 478 g 1 . 2 m 49 ◦ 611 g practice 09 – – Due: Mar 27 2007, 1:00 pm 2 9 . 8 m / s 2 After θ f At what speed will the lighter mass m 1 hit the heavier mass m 2 ? Correct answer: 8 . 29221 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 478 g , m 2 = 611 g , θ i = 49 ◦ , ℓ = 10 . 2 m , v i = before collision , V f = after collision , and g = 9 . 8 m / s 2 . The velocity just before the collision v i can be determined by energy conservation. When particle 1 is at its initial condition, it is at rest and displaced by an angle θ i from the vertical. The total energy is all potential and is given by U i = m 1 g ℓ (1- cos θ i ) where ℓ (1- cos θ i ) is the distance above the lowest point. Just before the collision, the en- ergy of sphere 1 is all kinetic energy, 1 2 m 1 v 2 1 i . Equating the two energies gives 1 2 m 1 v 2 1 i = m 1 g ℓ (1- cos θ i ) . Solving for V 1 i gives v 1 i = { 2 g ℓ [1- cos θ i ] } 1 / 2 = braceleftbig 2 (9 . 8 m / s 2 ) (10 . 2 m) × [1- cos(49 ◦ )] } 1 / 2 = 8 . 29221 m / s . Question 3, chap 10, sect 99. part 2 of 3 8 points After lighter sphere is let go and collides with the heavier sphere at the bottom of its swing, two spheres immediately bind to- gether. What is the speed of the combined system just after the collision? Correct answer: 3 . 63974 m / s (tolerance ± 1 %). Explanation: This is a perfectly inelastic collision. The speed of the two spheres after collision is de- termined by momentum conservation m 1 v 1 i = ( m 1 + m 2 ) V f (1) where m 1 is the mass and v 1 i is the initial velocity of sphere 1 just before the collision, m 2 is the mass of sphere 2, and V f is the velocity of the combined spheres just after...
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Kaplunovsky - practice 09 - practice 09 – – Due 1:00 pm...

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